What is the Maclaurin series of f(x) = cos(x)?

Nov 1, 2015

A Maclaurin series is simply the Taylor series centered around $a = 0$ specifically.

sum_(n=1)^(N) (f^((n))(0))/(n!) x^n
= (f(0))/(0!)x^0 + (f'(0))/(1!)x^1 +(f''(0))/(2!)x^2 + (f'''(0))/(3!)x^3 + ...

Thus, we need to take derivatives until we see a unique pattern.

${f}^{\left(0\right)} \left(x\right) = \textcolor{g r e e n}{f \left(x\right) = \cos x}$
$\textcolor{g r e e n}{f ' \left(x\right) = - \sin x}$
$\textcolor{g r e e n}{f ' ' \left(x\right) = - \cos x}$
$\textcolor{g r e e n}{f ' ' ' \left(x\right) = \sin x}$
$\textcolor{g r e e n}{f ' ' ' ' \left(x\right) = \cos x}$

So stopping at $n = 3$ is fine. We can still figure out the rest by the pattern. We get:

= (cos(0))/(0!)x^0 + cancel((-sin(0))/(1!)x^1)^(0) +(-cos(0))/(2!)x^2 + cancel((sin(0))/(3!)x^3)^(0) + ...

= 1/(0!) - x^2/(2!) + x^4/(4!) - x^6/(6!) + ...

$\textcolor{b l u e}{= 1 - {x}^{2} / 2 + {x}^{4} / \left(24\right) - {x}^{6} / \left(720\right) + \ldots}$